Divisibility of Spheres with Measurable Pieces

For an $r$-tuple $(\gamma_1,\ldots,\gamma_r)$ of special orthogonal $d\times d$ matrices, we say that the Euclidean $(d-1)$-dimensional sphere $S^{d-1}$ is $(\gamma_1,\ldots,\gamma_r)$-divisible if there is a subset $A\subseteq S^{d-1}$ such that its translations by the rotations $\gamma_1,\ldots,\gamma_r$ partition the sphere. Motivated by some old open questions of Mycielski and Wagon, we investigate the version of this notion where the set $A$ has to be measurable with respect to the spherical measure. Our main result shows that measurable divisibility is impossible for a"generic"(in various meanings) $r$-tuple of rotations. This is in stark contrast to the recent result of Conley, Marks and Unger which implies that, for every"generic"$r$-tuple, divisibility is possible with parts that have the property of Baire.


Introduction
Let SO( ) denote the group of special orthogonal × matrices, that is, real × matrices such that the determinant of is 1 and = , where denotes the identity × matrix.The elements of this group are naturally identified with orientation-preserving isometries of the Euclidean unit sphere and we will often refer to them as rotations.
Here we propose to study the more general question of describing the set of those -tuples γ ∈ SO( ) such that Ë −1 is γ-divisible with measurable pieces.
In other words, this property states that if a polynomial with rational (equivalently, integer) coefficients vanishes on (the matrix entries of) γ then it necessarily vanishes everywhere on SO( ) .
Our main result shows that no generic γ works in the measurable setting, even in a rather relaxed fractional version.
In sharp contrast, we can derive with some extra work from the results in [4] that every generic γ works with pieces that have the property of Baire.Proposition 1.2.Let 2 and 2 be arbitrary integers, except if is odd then we require that 3. Let ( 1 , ... , ) ∈ SO( ) be generic.Then there is a subset of Ë −1 with the property of Baire such that 1 . ,... , .partition Ë −1 .
Theorem 1.1 and Proposition 1.2 add to a growing body of results in measurable combinatorics (see e.g. the recent survey by Kechris and Marks [16]), where the requirements that the pieces are measurable and have the property of Baire respectively lead to different answers.
The following lemma shows that, in various meanings, "most" elements of SO( ) are generic.

Lemma 1.3. Let
1, 2 and N be the set of -tuples in SO( ) that are not generic.Then the following statements hold.
(i) The set N has measure 0 with respect to the Haar measure on the group SO( ) .
(ii) The set N is a meager subset of SO( ) with respect to the topology induced by the Euclidean topology on Ê 2 ⊇ SO( ) .
Also, by using some algebraic geometry, we can give a more concrete characterisation of generic -tuples of rotations.In particular, the following lemma allows us to write an "explicit" generic point: just let the entries above the diagonals be sufficiently small reals that are algebraically independent over É and extend this to an element of SO( ) by Claim 8.3 here.Lemma 1.4.Let 1, 2, and γ ∈ SO( ) .Then γ is generic if and only if the 2 -tuple of the matrix entries of γ strictly above the diagonals is algebraically independent over É.
Some standard general results of Borel combinatorics (e.g.Lemma 5.12 and Theorem 5.23 from [21]) imply that if Ë −1 is γ-divisible and every orbit of the subgroup of SO( ) generated by 1 , ... , is finite, then there is a Borel γ-division.The following result gives that just one finite orbit is enough to convert a γ-division into a measurable one.Proposition 1.6.Let 2 and γ = ( 1 , ... , ) ∈ SO( ) .Let be the subgroup of SO( ) generated by 1 , ... , .Suppose that there is z ∈ Ë −1 such that its - orbit .z is finite.Then Ë −1 is γ-divisible if and only if Ë −1 is γ-divisible with measurable pieces.
Of course, this leaves a wide range of unresolved cases.As an initial partial step, we completely characterise those -tuples of rotations for which the circle Ë 1 is divis- ible with measurable pieces for 3.
This paper is organised as follows.In Section 2 we give a quick overview of basic definitions and facts about spherical harmonics and use these to prove Theorem 1.1, which is the main result of this paper.Proposition 1.5 is proved in Section 3 using Euler's characteristic.Propositions 1.6 and 1.2 are proved in Sections 4 and 7 respectively.In Section 5 we describe the standard construction of how an -division of Ë −1 can be lifted to Ë +1 and observe that this gives measurable pieces (Lemma 5.1).In Section 6 we study various versions of measurable divisibility when = 2; in particular, we characterise -tuples γ ∈ SO(2) for which the circle Ë 1 is γ-divisible with measurable pieces for 3.The rather technical Section 8 is dedicated to proving Lemmas 1.3 and 1.4.Section 8.1 presents some basics of algebraic geometry.In Section 8.2 we prove some results about SO( ) and use them to prove Lemma 1.3.
In particular, we show that the variety SO( ) ⊆ Ê 2 is irreducible and the entries above the diagonals form a transcendence basis for its function field.While these results are fairly standard, we present their proofs since we could not find any published statements that suffice for our purposes.In Section 8.3 we prove an auxiliary lemma from algebraic geometry and use it to derive Lemma 1.4.

Let an integer
2 be fixed throughout this section.
For an introduction to spherical harmonics on Ë −1 we refer to the book by Groe- mer [10] whose notation we generally follow.Recall that denotes the spherical measure on Ë −1 .Thus the total measure of the sphere is As is fixed, the dependence on is usually not mentioned except for (since −1 will also appear in some formulas).Also, the shorthand a.e.stands for -almost everywhere.
By [10,Lemma 1.3.1], the density of the push-forward of under the projection to any coordinate axis is is the Laplace operator.A spherical harmonic is a function from Ë −1 to the reals which is the restriction to Ë −1 of a harmonic polynomial on Ê .Let H be the vector space of all spherical harmonics.For an integer 0, let H ⊆ H be the linear subspace consisting of all functions : Ë −1 → Ê that are the restrictions to Ë −1 of some harmonic polynomial which is homogeneous of degree , where we regard the zero polynomial as homogeneous of any degree.By [10,Lemma 3.1.3],the polynomial is uniquely determined by ∈ H , so we may switch between these two where we agree that + −3 −2 = 0 for = 0 or 1.Let •, • denote the scalar product on 2 (Ë −1 , ) (while x • y := =1 denotes the scalar product of x, y ∈ Ê ).It is known ( [10, Each space H is invariant under this action ([10, Proposition 3.2.4])since, on Ê , rotations preserve both the Laplace operator as well as the set of homogeneous degreepolynomials.
An important role is played by the Gegenbauer polynomials ( 0 , 1 , ... ) which are obtained from (1, , 2 , ... ) by the Gram-Schmidt orthonormalization process on 2 ([−1, 1], ( ) d ), except they are normalised to assume value 1 at = 1 (instead of being unit vectors in the 2 -norm).In the special case = 3 (when is the constant function), we get the Legendre polynomials.Of course, the degree of is exactly .Let us collect some of their standard properties that we will use.

Lemma 2.1. For every integer
0 the following holds.(i) The polynomial has rational coefficients.
(ii) For every v ∈ Ë −1 , the function v : Ë −1 → Ê, defined by form a basis of the vector space H .
Part (ii), namely the claim that each v is in H , is one of the statements of [10,Theorem 3.3.3].
Part (iii) is the content of [10,Theorem 3.3.14].Alternatively, notice that under the action in (2.4) we have for every u, v ∈ Ë −1 and ∈ SO( Thus the linear span of v , v ∈ Ë −1 , is a non-zero SO( )-invariant subspace of H .
By [10,Theorem 3.3.4],the only such subspace is H itself, giving the required.Part (iv) follows from where the first equality is a special case of the Funk-Hecke Formula ([10, Theorem 3.4.1])and the second equality (which by (2.1) amounts to computing the 2 -norm of any u ∈ 2 (Ë −1 , )) is proved in [10,Proposition 3.3.6].
We need the following strengthening of Lemma 2.1.(iii),where we additionally require that the vectors v are rational.Lemma 2.2.For every integer 0, there is a choice of v 1 , ... , v ∈ Ë −1 ∩ É such that the functions v , ∈ [ ], form a basis of the vector space H .
Proof.We pick v in Ë −1 ∩ É one by one as long as possible so that the corres- ponding functions v are linearly independent as elements of H . Let this procedure produce v 1 , ... , v ℓ .Suppose that ℓ < as otherwise we are done.Let v ℓ+1 = x, with x = ( 1 , ... , ) ∈ Ë −1 being viewed as a vector of unknown variables.Consider the In other words, is the Gram matrix of the vectors v 1 , ... , v ℓ+1 ∈ 2 (Ë −1 , ).In particular, the determinant det( ) of is 0 if and only if v ℓ+1 is in the span of the (linearly independent) vectors v 1 , ... , v ℓ (by e.g.[14, Theorem 7.2.10]).
By Lemma 2.1.(iv)we have that Thus the determinant of is a polynomial function of x.By Lemma 2.1.(iii)and ℓ < (and the linear independence of v 1 , ... , v ℓ ), there is some choice of v ℓ+1 ∈ Ë −1 with det( ) ≠ 0. That is, the polynomial det( ) is not identically zero on Ë −1 .
We need the following easy claim that can be proved, for example, by induction on 2 with the base case = 2 following from Ë 1 containing all points of the form Claim 2.3.For every 1, the set Ë −1 ∩ É of the points on the sphere with all coordinates rational is dense in Ë −1 with respect to the standard topology on the sphere (i.e. the one inherited from the Euclidean space Ê ⊇ Ë −1 ).
Since det( ), as a polynomial function of x ∈ Ë −1 , is continuous and not identic- ally zero, it has to be non-zero on some point x of the dense subset Ë −1 ∩ É .Thus, if we let v ℓ+1 to be such a vector x, then the functions v 1 , ... , v ℓ+1 ∈ 2 (Ë −1 , ) are linearly independent.This contradiction to the maximality of v 1 , ... , v ℓ proves the lemma.
For an integer 0, an -tuple γ = ( 1 , ... , ) ∈ SO( ) and a unit vector By Lemma 2.1.(ii),each function v ,γ : Ë −1 → Ê, as a linear combination of some spherical harmonics Proof.By Lemma 2.2, we can fix some vectors v 1 , ... , v ∈ Ë −1 ∩ É such that v 1 , ... , v form a basis for H . Let β = ( 1 , ... , ) be an arbitrary element of SO( ) (not necessarily generic).Consider the × matrix = (β) with entries Recall that the vectors v , ∈ [ ], form a (not necessarily orthonormal) basis of the linear space H . Write the vectors v ,β in this basis: for some × matrix .Then is the matrix product , where is the Gram matrix of the vectors v multiplied by the constant −1 (that is, the entries of are defined by the formula in (2. Since v 1 , ... , v are fixed, this writes each as a polynomial in the 2 entries of the matrices 1 , ... , .Moreover, all coefficients of this polynomial are rational since each v belongs to É and all coefficients of are rational by Lemma 2.1.(i).
Thus the determinant of is equal to (β) for some polynomial with coefficients in É.
Note that if we let each be the identity matrix , then v ,β becomes v for every v ∈ Ë −1 and we have = v , v for , ∈ [ ] and det( ) ≠ 0 (since v 1 , ... , v are linearly independent).Thus ( , ... , ) ≠ 0. Since γ ∈ SO( ) is generic, we have that (γ) ≠ 0, that is, the matrix for β := γ is non-singular.This means that the functions v ,γ , ∈ [ ], are linearly independent.Since they all lie in H and their number equals the dimension of this linear space, they span H .The lemma is proved.
Given the above auxiliary results, we can derive Theorem 1.1 rather easily.
Remark 2.5.The statement of Theorem 1.1 remains true also when = and ( 1 , ... , −1 ) is a generic point of SO( ) −1 .One way to see this is to run the same proof except the -th component of each encountered -tuple of matrices is always set to be the identity matrix .

Rotations generating a finite subgroup
Proof of Proposition 1.5.We have to show that an even-dimensional sphere Ë −1 is not ( 1 , ... , )-divisible if the subgroup of SO( ) generated by the rotations 1 , ... , is finite.
Since is odd, the 2-divisibility of Ë −1 is impossible because of a fixed point of −1 1 2 .So assume that 3. Let := .{±e 1 , ... , ±e }, that is, we take all possible images of the standard basis vectors and their negations when moved by .Clearly, the set is a finite.Let be the convex hull of .Then is a full-dimensional polytope containing 0 in its interior (as already the convex hull of {±e 1 , ... , ±e } ⊆ has these properties).Its boundary is homeomorphic to Let a hyperplane mean a ( − 1)-dimensional affine subspace of Ê .Identify each oriented hyperplane ⊆ Ê with the pair (n, ) ∈ Ë −1 × Ê so that

Its open half-spaces are
Call supporting if ∩ ≠ ∅ and − ∩ = ∅.Call a facet hyperplane if it is supporting and dim aff ( ∩ ) = − 1, where dim aff ( ) denotes the dimension of the affine subspace of Ê spanned by .
The intersections of supporting hyperplanes with represent the boundary of the polytope as a CW-complex.Namely, for ∈ {0, ... , − 1}, its -dimensional cells are precisely the -dimensional faces of , that is, the convex hulls of the sets in Let us show that for every ∈ {0, ... , − 1} and distinct , ∈ C we have m ≠ m .As it is well-known, see e.g.[11,Theorem 3.1.7],we can pick facet hyperplanes 1 , ... , such that ∩ (∩ =1 ) = .Since ≠ , the affine subspaces that these two sets span differ.Since these subspaces have the same dimension, there is y ∈ not in the affine span of .Since y ∈ and each is supporting, there is ∈ [ ] such that y belongs to the open half-space + .From ⊆ ∪ + , it follows that m belongs to + and cannot be equal to m ∈ , as claimed.
Also, it holds that m ≠ 0 for any ∈ C .Indeed, with 1 , ... , as above we have that 0, which is in the interior of , belongs to, say, the open half-space + 1 so cannot be equal to m ∈ 1 .
Remark 3.1.Under the assumptions of Proposition 1.5, its proof gives that if there are linearly independent vectors on Ë −1 such that each has a finite orbit under (where some of these orbits may coincide) then Ë −1 is not γ-divisible.However, this seemingly weaker assumption is equivalent to the assumption that is finite (e.g. via a version of Claim 4.2 below).

Actions with a finite orbit
Here we prove Proposition 1.6 that, in the presence of at least one finite orbit, γ-divisibility is equivalent to measurable γ-divisibility.
For x ∈ Ë −1 , let x be the linear subspace of Ê spanned by .x⊆ Ê .Claim 4.1.For every x ∈ Ë , both x ⊆ Ê and its orthogonal complement ⊥ x ⊆ Ê are invariant under the action of on Ê .Proof of Claim.Any ∈ permutes the set .x.Since is a linear map, it preserves the linear subspace x spanned by .x.Thus x is -invariant.
Since consists of orthogonal matrices, its action preserves the scalar product on Ê .Thus if y ∈ Ê is orthogonal to x then, for every ∈ , we have that .y Recall that z ∈ Ë −1 is a vector such that its orbit .z is finite.Let 1 , ... , be the elements of .z.
Fix ⊆ Ë −1 such that 1 ., ... , .partition Ë −1 .By the invariance of z and ⊥ z , the translates of the set ∩ (Ê × 0) (resp.∩ (0 × Ê − )) by 1 , ... , partition Ë −1 × 0 (resp. 0× Ë − −1 ).By Claim 4.2, every orbit of the action of on the invariant subset := Ë −1 × 0 has at most !elements.Obviously, the same holds for the action on Ë −1 of the subgroup ′ ⊆ O( ) generated by 1 , ... , .Fix a Borel total order on Ë −1 (e.g. the restriction of the lexicographic order on Ê ) and let ′ ⊆ be obtained by picking from every orbit ′ .x⊆ Ë −1 the lexicographically smallest subset such that its translates by 1 , ... , partition ′ .x.Such a set always exists since {y ∈ ′ .x| (y, 0) ∈ } is one possible choice.In the terminology of [21], the set ′ can be computed by a local rule of radius ! on the coloured Schreier digraph of ′ Ë −1 (where the vertex set is Ë −1 and we put a directed colour-arc from y to .y for all y ∈ Ë −1 and ∈ [ ]).As the action is Borel, this is known to imply (see e.g.[21, Lemma 5.17]) that the constructed set ′ ⊆ Ë −1 is Borel.Define and, as we observed earlier, 1 ., ... , .partition 0 × Ë − −1 .Thus ∪ witnesses the γ-divisibility of Ë −1 .Note that the set , which lies inside the intersection of Ë −1 with the linear subspace ⊥ z of dimension less than , has measure zero.On the other hand, the set can be equivalently defined as the pre-image of the Borel set ′ × Ê − under the natural homeomorphism between Ë −1 \ (0 × Ë − −1 ) and Ë −1 × Ê − that maps (x, y) to (x/ x 2 , y/ x 2 ).Thus is Borel and ∪ is measurable, proving the proposition.⊆ SO( ) has a finite orbit of size at least 3, then is finite (and thus Proposition 1.5 applies).However, this implication is not true in general for 4. For example, we can take the subgroup of SO( ) generated by a diagonal block matrix whose first (resp.second) block is a 2 × 2 special orthogonal matrix of order 3 (resp.of infinite order), while all remaining blocks are the 1 × 1 identity matrices.Then has an infinite order (coming from the second block) but its action on Ë −1 has an orbit with exactly 3 elements (e.g. the orbit of the first standard basis vector (1, 0, ... , 0)).

Measurable divisibility of higher-dimensional spheres
As we mentioned in the Introduction, Ë −1 is -divisible with Borel pieces for every 2 and even 2 ([23, Theorem 6.6(a)]).The proof of [23, Theorem 6.6(b)] for any 3 and odd 5 gives measurable pieces.Since this conclusion does not seem to be explicitly stated anywhere in [23], we provide the simple proof from [23].
Proof.Informally speaking, we will use the Borel -divisibility of Ë 1 in the last two coordinates of Ë −1 ⊆ Ê , resorting to the -divisibility of Ë −3 only on the null set of points where the last two coordinates are zero.Namely, choose rotations 1 , ... , ∈ SO( − 2) and a (not necessarily measurable) subset ⊆ Ë −3 such that 1 . ,... , .partition Ë −3 , which is possible by e.g.[23, Theorem 6.6].Let ∈ SO(2) be the rotation of the circle Ë 1 by the angle 2 / .(Thus the order of , as an element of the group SO(2), is .)For ∈ [ ], let send (x, y) ∈ Ê −2 × Ê 2 to ( .x,.y),where we view SO( ) as also acting on Ê .Clearly, preserves both the scalar product on Ê and the orientation; thus it is an element of SO( ).

Measurable divisibility for = 2 and 4
We parametrise Ë 1 = {(cos , sin ) | ∈ [0, 2 )} and use the parameter instead of the Cartesian coordinates.Thus we have the interval [0, 2 ) with being the Lebesgue measure on it.The space H for 1 becomes the span of cos and sin (while, of course, H 0 consists of all constant functions).Here, the harmonic expansion is nothing else as the Fourier series.We identify SO(2) with the additive group Ì := Ê/2 of reals taken modulo 2 .Thus the action of ∈ Ì on [0, 2 ) is to send ∈ [0, 2 ) to + (mod 2 ).We also identity [0, 2 ) with Ì; thus we have the natural action Ì Ì.
If differs from 1/ on a set of positive measure then, for at least one integer 1, we have ( , ) ≠ (0, 0) and thus (6.1) holds.Conversely, if (6.1) holds for some 1, then we can take, for example, ( ) This completely describes the set of -tuples in SO(2) for which the circle Ë 1 is "fractionally" divisible: Proposition 6.1.An -tuple ( 1 , ... , ) ∈ Ì belongs to F if and only if (6.1) holds for at least one integer 1.
Let us investigate the sets B and M ′ for 4. As we will see, it holds for each 4 that B = M ′ (and, in particular, this set is also equal to M ).
Proof of Claim.The non-trivial direction of the claim can be derived by observing that, up to a permutation of indices, we can assume that v := ( 1 , 1 ) + ( 2 , 2 ) is a non-zero vector while, in general, there is at most one way to write −v ∈ Ê 2 \ {0} as the unordered sum of two unit vectors.Thus the other two vectors must be (− 1 , − 1 ) and (− 2 , − 2 ), as desired.
This implies that t ∉ M ′ 4 .Unfortunately, an explicit characterization of the set B 4 = M 4 = M ′ 4 for general seems to be rather messy, although it reduces to a finite case analysis for any given t ∈ Ì 4 by Claim 6.3.So we will restrict ourselves to the special cases = 1 and = 2, just to illustrate that the measurable t-divisibility is not determined by the order 4 of the group alone (which happens already for = 2).
First, suppose that an -set ⊆ 4 witnesses the k-divisibility.Since is odd, some residue modulo appears an odd number of times in .This multiplicity cannot be larger than 2 since otherwise the translates 1 + , ... , 4 + would cover the four points , + , + 2 , + 3 ∈ 4 at least six times.Thus the multiplicity of in modulo is exactly 1.By the commutativity of 4 , we can replace by any its translate.Thus assume that contains 0 but none of , 2 and 3 .Thus, by ( 3 , 4 ) = ( , 0), the set ( 3 + ) ∪ ( 4 + ) covers 0 and but not 2 nor 3 .Since 2 ∉ , the only way to consistently cover 2 and 3 is that 2 − ∈ .Now, { 1 , ... , 4 } + {0, 2 − } contains 2 − and 3 − but not − nor − .None of the last two elements can be covered by 3 + or 4 + (as then modulo would contain − (mod ) at least twice but then the four elements 0, , 2 , 3 would be covered at least six times, with the extra multiplicity coming from 0 and being covered by 0 ∈ when translated by 3 and 4 ).Thus the only way to consistently cover − and − is that −2 ∈ .One can continue to argue in this manner, showing that for each ∈ {0, 1, ... } we have −2 ∈ and −(2 + 1) + 2 ∈ .As the first of these elements of are pairwise distinct (in fact, they have pairwise distinct residues modulo ) and is odd, it must hold that the -th element, − + 2 , belongs to .Since does not contain any of , 2 and 3 , we necessarily have that − + 2 ≡ 0 (mod 4 ).This equation has solutions, namely, all ∈ 4 with ≡ 2 (mod 4), giving the claim.
In the initial version of the manuscript, we conjectured that if ( 1 , ... , ) ∈ M ′ then ( − )/ is rational for every , ∈ [ ].This conjecture was subsequently proved by Grebík, Greenfeld, Rozhoň and Tao [9].This implies that B = M = M ′ for every (by an argument similar to that of Proposition 1.6) and reduces the question if any given t ∈ Ì belongs to this set to some finite case analysis.

Proof of Proposition 1.2
In order to prove Proposition 1.2, we need some auxiliary results first.Proof.If some non-zero x ∈ Ê satisfies x = 0 for every ∈ [ ], then x = =1 ( x) = 0, so the determinant of is zero.Conversely, suppose that is singular.Choose a non-zero vector x ∈ Ê with and each x must be the zero vector, giving the required.
The results of Dekker [6], Deligne and Sullivan [8], and Borel [3] (see Theorem 6.4 in [23] and the historical discussion preceding it) give the following.

Lemma 7.2. For every
2 and 2 there is a choice of rotations 1 , ... , ∈ SO( ) that generate the free rank-group such that its action on Ë −1 is free for even and locally commutative for odd (meaning that every two elements of that have a common fixed element on Ë −1 commute).
Note that the above result is usually stated in the special case = 2 as the general case easily follows by taking any subgroup of 2 isomorphic to .Proof.For a non-trivial reduced word in and β = ( 1 , ... , ) ∈ SO( ) , the relation (β) = amounts to 2 polynomial equations, with (β) = 0 stating that the ( , )-th entry of the corresponding product of the matrices of 's and their transposes (which are equal to their inverses) is ½ = , where ½ = is 1 if = and 0 otherwise.Each of these polynomials has rational coefficients.Moreover, the -tuple of matrices β returned by Lemma 7.2 (which, in particular, generates the free subgroup) gives a point where at least one of these polynomials is non-zero, say (β) ≠ 0. The polynomial has to be non-zero also at the generic point γ ∈ SO( ) and so (γ) ≠ .Since was an arbitrary non-trivial word, the rotations 1 , ... , indeed generate the free group.
Let us show the second part in the case of odd (with the case of even being similar).Suppose on the contrary that we have two reduced non-commuting words 1 and 2 in such that the corresponding elements 1 (γ) and 2 (γ) have a common fixed point x ∈ Ë −1 .Thus the matrices 1 := 1 (γ) − and 2 := 2 (γ) − have x ≠ 0 as a common zero eigenvector.By Lemma 7.1, this property is equivalent to det( 1 1 + 2 2 ) = 0, which is a polynomial equation in γ with rational coefficients.For the special -tuple of matrices β returned by Lemma 7.2, the matrices 1 := 1 (β) − and 2 := 2 (β) − cannot have a common zero eigenvector as it would give a common fixed point for the non-commuting elements 1 (β) and 2 (β).Thus, we have by Lemma 7.1 that det( 1 1 + 2 2 ) ≠ 0. We have found a polynomial equality with rational coefficients that holds for γ but not for β ∈ SO( ) .This contradicts our assumptions that γ ∈ SO( ) is generic.Also, we will need the following result of Conley, Marks and Unger that directly follows (as a rather special case) from Lemmas 3.4 and 3.6 in [4].
For every ∈ SO( ) \ { }, the set of its fixed points on Ë −1 is closed (as the preimage of 0 under the continuous map that sends x ∈ Ë −1 to .x− x ∈ Ê ) and has empty relative interior (for otherwise one can choose linearly independent vectors fixed by , contradicting ≠ ).In particular, this set is meager.Since the group is countable, the free part of the action (which consists of x ∈ Ë −1 such that .x≠ x for each non-trivial ∈ ) is co-meager.Also, it is easy to show that the free part is a Borel subset of the sphere (see e.g.[21,Lemma 4.4]).

Proof of Lemmas 1.3 and 1.4
This section is dedicated to proving Lemmas 1.3 and 1.4.Their proofs are rather technical; this is why we postponed them until the very end.

Some definitions and results from algebraic geometry.
In this section we present some definitions and results from algebraic geometry that we need.We will follow the notation from the book by Hassett [12] to which we refer for missing details (and for a nice concrete introduction to most results needed here).
A field extension ↩→ is called algebraic if every ∈ is algebraic over , that is, satisfies a non-trivial polynomial equation with coefficients in .Some easy but very useful facts ([12, Proposition A.16]) are that, for an arbitrary field extension ↩→ , (8.1) the elements of that are algebraic over form a field and, for another field extension ↩→ , ( if ↩→ and ↩→ are both algebraic then ↩→ is algebraic.
Let us fix a field .By a variety we mean a subset of some affine space which is closed in the Zariski topology, that is, is equal to for some family F ⊆ [x] of polynomials where x := ( 1 , ... , ).Then the coordinate ring of is denotes the ideal of the variety ⊆ .We call a variety ⊆ irreducible if we cannot write = 1 ∪ 2 for some varieties 1 , 2 . This is equivalent to the statement that the ideal ( ) ⊆ [x] is prime ([12, Theorem 6.5]).Then [ ] is a domain so we can define its fraction field, which is called the function field of and is denoted by ( ).Elements of [ ] (resp.( )) can be viewed as the restrictions of polynomial (resp.rational) functions to modulo identifying functions that coincide on .
We will also need the following easy result.

Variety SO( ;
) .In this section we show in particular that SO( ) , as a variety in Ê 2 , is irreducible and that the set of entries above the diagonals forms a transcendence basis; in particular, the dimension of SO( ) is 2 .In fact, we will need an extension of this result, where the underlying field can be different from Ê, for the proof of Lemma 1.4 (even though the statement of Lemma 1.4 deals only with the real case).
Let 1 be an integer and be a field.Consider the affine space × of all × matrices with entries in , writing its elements as = ( , ) , ∈ [ ] .Let the special orthogonal variety over be the variety SO( ; ) := ( SO ) ⊆ × defined by the ideal (8.4) where := 2 ,1 + ... + 2 , − 1 encodes the fact that each row is a unit vector (when ⊆ Ê), , := ,1 ,1 + ... + , , encodes the orthogonality of the -th andth rows while the last constraint states that the determinant of is 1.Note that the "orthonormality" constraints force to have determinant −1 or 1, which follows from ).
The matrix multiplication makes SO( ; ) a group.If = Ê then we get the familiar group SO( ) of special orthogonal real × matrices (and the shorthand SO( ) will always be reserved for the real variety SO( ; Ê)).
be the sequence of the 2 entries strictly above the diagonals.We call these entries upper.For notational convenience, we fix an ordering of the coordinates of 2 so that all non-upper entries (that is, those on or below the diagonals) come before all upper ones; thus when we write a vector of length 2 as (x, y) then we mean that y is the upper part.
Lemma 8.2.For every subfield ⊆ , the variety := (SO( ; )) ⊆ 2 is irreducible, has dimension 2 and the set of upper coordinates forms a transcendence basis of the function field ( ) over .
Proof.First, let us show that is irreducible The proof of this in the case = 1 (for an arbitrary field with 2 ≠ 0) can be found in [2, Proposition 5-2.3].We adopt the argument from [2] to work for any 1.(Note that products need not preserve the irreducibility when the underlying field is not algebraically closed.)For x ∈ with x • x := =1 2 non-zero, the map x : → that is defined by x, for y ∈ , can be thought of as the reflection of around the hyperplane orthogonal to x, so we call x a reflection.Each ∈ SO( ; ) can be written as a product of an even number of reflections, see [2, Proposition 1-9.4] (and, conversely, every such product is in SO( ; )).In fact, the proof in [2], which proceeds by induction on , shows that at most := 2 reflections are needed.By inserting the trivial composition x x = for some x ∈ with x • x ≠ 0 we can write each ∈ SO( ; ) as the product of exactly reflections.
Consider the product map : ( ) → SO( ; ) that applies in each of the coordinates.As the complement := \ is Zariski closed (as the finite union over ∈ [ ] of the sets of (z 1 , ... , z ) ∈ satisfying the polynomial equation z • z = 0), the complement := \ is also Zariski closed as the finite union over ∈ [ ] of the closed sets ( −1) × × ( − ) .Clearly, is a rational map defined everywhere on and thus continuous in the Zariski topology on ⊆ .Also, the image of is exactly = SO( ; ) with the surjectivity following from the choice of .It follows from [2, Lemma 5-2.1] that is irreducible.(In brief, if can be written as a union of two proper closed subsets 1 ∪ 2 , then is a union of two proper closed sets −1 ( 1 ) ∪ and −1 ( 2 ) ∪ , contradicting the irreducibility of since its ideal ( ), which is {0} by e.g.Lemma 8.1, is trivially prime.)Thus is indeed irreducible.
It remains to show that the set of upper coordinates (that is, all entries above the diagonals) is a transcendence basis for the function field ( ) over .This claim is made of the following two parts.
First, let us show that the field extension ( ) ↩→ ( ) is algebraic.By (8.1) and (8.2), it is enough to represent this field extension as a composition of field extensions where, at each step, every added non-upper coordinate is algebraic over the previously added coordinates and the upper coordinates in the same matrix.Thus we consider just one matrix in SO( ; ), which we denote as = ( , ) , ∈ [ ] .We add the non-upper coordinates by whole rows in the natural order (with Row 1 added first, then Row 2, and so on).Take any Row and a non-upper pair ( , ) (i.e. with ).The following argument works for every index ∈ [ ] so we pick = for notational convenience.Thus we have to show that := , , as an element of ( ), is algebraic over Let the vector x := ( ,1 , ... , , −1 ) consist of the other non-upper entries of Row and let := ( , ) , ∈ [ −1] be the square submatrix of which lies above x.The orthogonality of Row to the previous rows gives a system of − 1 linear equations, namely, x = f , holds.Moreover, if , for each ( , ) ∈ is real then ,1 , ... , , can additionally be chosen to be real.
Proof of Claim.Suppose that the claim fails for for some ∈ [ ] and > 0. Let real tend to 0 from above and let ∈ be a partial assignment violating the claim.Let us use the notation that was introduced around (8.6).By our choice of , we have that each entry of is within additive = (1) from the corresponding entry of the identity matrix and thus det( ) = 1 + (1) is non-zero.Of the two roots of the quadratic equation (8.6), which now reads In fact, (8.6) gives not only the entry = , but the consistent remainder of Row by x := (det( )) −1 Ad( )f , satisfying P .By the continuity of the all involved functions (and det( ) = 1 + (1)), we have x ∞ = (1), a contradiction to > 0 being fixed.
Let us show how to adapt this argument to establish the second part of the claim.Suppose additionally that the given , 's are reals.In the above notation, the quadratic equation (8.6) has all real coefficients and, as before, states that 2 − 1 = (1 + | 2 |).Its left-hand side as a function of ∈ Ê changes sign at = 1 with its derivative 2 being bounded away from 0 around = 1.Hence we can choose a real root = 1 + (1).Then is a real matrix and the rest of Row , namely x := (det( )) −1 Ad( )f is also real.
Consider the projection : SO( ; ) → on the := 2 upper coordinates, which maps (x, y) to y.In particular, the -tuple of the identity matrices projects to the zero vector 0 ∈ . The image of contains some Euclidean open ball Ball (0) := {z ∈ | z 1 < } of radius > 0 around the origin.Namely, we can take its radius to be (8.7):= ( −1 ( ... 1 (1/(2 !)) ... )) > 0, where 1 , ... , are the functions returned by Claim 8.3.Indeed, by the choice of the constants we know that for every y ∈ Ball (0), we can construct a × matrix row by row so that projects to y and satisfies all properties P 1 , ... , P while it also holds that − ∞ < 1/(2 !).The last inequality gives, rather roughly, that | det( ) − 1| < 1.Thus det( ) = 1 because det( ) is either −1 or 1 by (8.5).So indeed (SO( ; )) contains ( ) = y.Now, suppose on the contrary that there is a non-trivial polynomial relation between the upper coordinates.Thus there is a non-zero polynomial which does not depend on the non-upper coordinates and belongs to the ideal generated by the polynomials that define SO( ; ) (with those for = 1 being listed in (8.4)).The polynomial , as a function of the upper coordinates, vanishes on ( ) ⊆ .This Since each polynomial is continuous as a function Ê 2 → Ê, each set is closed.
Let us turn to Part (i) where we have to show that the Haar measure assigns measure 0 to N .By the countable additivity, it is enough to show that each set , defined by (8.8), has -measure zero.
The set SO( ) , as an algebraic variety, has dimension smaller than which follows from the definition of the dimension via nested chains of irreducible varieties (that is, by (8.3)) and from the irreducibility of the variety SO( ) (that is, by Lemma 8.2).Some standard results in the theory of (semi-)algebraic sets give that every bounded variety in some Ê admits a triangulation into simplices each of which is a smooth submanifold of Ê , see e.g. [1,Theorem 5.43].Apply this result to every irreducible component ⊆ .The dimension of each obtained simplex (as a manifold) is at most dim .Indeed, pick a point s ∈ and the projection from on some coordinates which is a homeomorphism around s. Observe that these coordinates are algebraically independent in the function field Ê( ) because no non-zero polynomial on Ê can vanish on a non-empty open set by Lemma 8.1.
Thus we covered by finitely many manifolds of dimension less than , each having zero Haar measure as it was observed earlier (by [17,Equation (8.25)]).We conclude that the Haar measure of is indeed zero.
Let us show Part (ii).Recall that the sets 1 , 2 , ... were defined in (8.8).Clearly, each set is closed.Thus it is enough to show that the relative interior of each ⊆ SO( ) is empty.Suppose on the contrary that the relative interior of some is non-empty.Since the compact group SO( ) acts transitively on itself by homeomorphisms, finitely many translates of cover the whole group.As the Haar measure is is invariant under this action, we have that ( ) > 0. However, this contradicts Part (i) that we have already proved.
This finishes the proof of Lemma 1.3.

Proof of Lemma 1.4.
Our proof of the reverse (harder) implication of Lemma 1.4 needs Lemma 8.4 below.Since we could not find this rather natural statement anywhere in the literature we present a proof whose main idea (to use dimension) was suggested to us by Miles Reid.In fact, Miles Reid came up with a full proof of some initial version of the lemma.Since his proof relies on the so-called universal domain of while we would like to have this paper as elementary as possible, we present a proof that avoids universal domains.Given a field extension ↩→ and a variety ⊆ (over the field ), we say that an element a ∈ is -generic for if every polynomial ∈ [ 1 , ... , ] with (a) = 0 vanishes on every element of .(Here as well as in the rest of this paper, each evaluation mixing elements of some two fields ↩→ is done in the larger field .)In the special case when := É, := Ê, := SO( ) we get exactly the definition of a generic -tuple of rotations from the Introduction.As P ⊆ , we trivially have that ⊆ .We have to show that = , which by the definition of = ( ) will give the required result (namely, that every ∈ vanishes on ).Let = 1 ∪ ... ∪ be a decomposition of into irreducible varieties ([12, Theorem 6.4]).
Suppose first that there is ∈ [ ] such that the -tuple y, with each viewed as an element of the function field ( ), is algebraically independent over .This means that the dimension of the irreducible variety ⊆ + is at least .Recall that ⊆ ⊆ .By the definition of the dimension via nested chains of irreducible subvarieties (that is, by (8.3)), we cannot have for otherwise any chain for extends to a strictly larger chain for which gives that dim − 1 dim , contradicting our assumption.Thus = = , as desired.
Thus | | = |C |, where := {m / m 2 | ∈ C } ⊆ Ë −1 denotes the set of the normalised centres of mass of the vertex sets of -dimensional faces.Clearly, the set family C is invariant under the natural action of on finite subsets of Ë −1 .Thus the set Now we are ready to prove the (non-trivial) forward direction of Proposition 1.6.By rotating the sphere (and moving z and conjugating 's accordingly), we can assume that z = Ê × 0 and ⊥ z = 0 × Ê − for some ∈ [ ].By Claim 4.1, every matrix , ∈ [ ], consists now of two diagonal blocks that correspond to some ∈ O( ) and ∈ O( − ).(Note that these matrices may have determinant −1.)

Remark 4 . 3 .
One can show via Claims 4.1 and 4.2 that if = 3 and a subgroup
can combine the Borel γ-division of Ë −1 \ given by Conley, Marks
[4]pose that 1 / is rational.Let be the subgroup of Ì that is generated by 1 , 2 and 3 .(Thereisnoneedto add 4 as it is 0.) By (6.2) and the rationality of 1 / , the group is finite.Of course, if 4 does not divide its order | | then there is no t-division even if a null set can be removed.So suppose that | | = 4 for some integer , i.e. that is the cyclic group of order 4 .For ∈[4], let ∈ {0, ... , 4 − 1} satisfy that = 2 .Let k := ( 1 , ... , 4 ).Let us say that the cyclic group 4 , that consists of integer residues modulo 4 , is k-divisible if there is a subset ⊆ 4 such that the sets + , ∈[4], partition 4 .Of course, such a set must have exactly elements.
contradicts Lemma 8.1 as ( ) contains a non-empty open set, namely the open ball of radius around the origin, and thus ( ) contains a product of infinite sets.Now we are ready to show that the set N of non-generic points in SO( ) is "small".Proof of Lemma 1.3.As before, when we identify an -tuple of × matrices over a field with an element of 2 , let us order the 2 coordinates so that the := 2 upper entries (i.e.those above the diagonals) come at the end.Thus if we write an element of 2 as (x, y) then y corresponds to the upper entries.Also, we use the standard topology on Ë −1 (the one which is inherited from the Euclidean space Ê ).There are countably many polynomials in É[x, y] so enumerate those that are non-zero on at least one element of SO( ) as 1 , 2 , ... .By definition, if a point (a, b) ∈ SO( ; Ê) is not generic then some vanishes on (a, b).Thus N is a